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 Taylor polynomial. System of equations for the coefficients. marraco Fellow Posts: 93 Threads: 11 Joined: Apr 2011 04/30/2015, 03:24 AM (This post was last modified: 05/07/2015, 07:25 AM by marraco.) This is an attempt to get the system of equations for the calculation of the coefficients aᵢ of the Taylor polynomial developed around 0. We want the coefficients aᵢ of this Taylor expansion: ${^xa}=\sum_{n=0}^{\infty}{a_n .x^n}$ They should match this equation: ${\color{Blue} {^{x+1}a}$$\,=\,$${\color{Red} a^{^xa}}$ Let's start with the left side. We compose the Taylor expansion (on the first equation) with (x+1): ${\color{Blue} {{^{(x+1)}a}=\sum_{n=0}^{\infty}{a_n.(x+1)^n}}$ We use the binomial theorem $ \left.\begin{matrix} (1+x)^n = \sum_{k=0}^n {n \choose k}x^k \\ {n \choose k} = \frac{n!}{k!\,(n-k)!} \end{matrix}\right\} {^{(x+1)}a}=\sum_{n=0}^{\infty}{a_n.\sum_{k=0}^n {n \choose k}x^k}$ Each term on the summation is $ a_0.(1+x)^0 = a_0. \left({0 \choose 0}x^0 \right)\\ a_1.(1+x)^1 = a_1. \left({1 \choose 0}x^0 + {1 \choose 1}x^1\right)\\ a_2.(1+x)^2 = a_2. \left({2 \choose 0}x^0 + {2 \choose 1}x^1 + {2 \choose 2}x^2\right)\\ \vdots \\ a_n.(1+x)^n = a_n. \left({n \choose 0}x^0 + {n \choose 1}x^1 + {n \choose 2}x^2 + \cdots + {n \choose {n-1}}x^{n-1} + {n \choose n}x^n\right)$ In resume: ${\color{Blue} {{^{(x+1)}a}=\sum_{n=0}^{\infty}\left({\sum_{m=n}^{\infty}{{m \choose n}.a_m}\right).x^n$ $^{x+1}a=( a_{0} + a_{1} + a_{2} + a_{3} + a_{4} + a_{5} + a_{6} + a_{7} + a_{8} +a_{9} + a_{10} + ...)+\\ (a_{1} + 2a_{2} + 3a_{3} + 4a_{4} + 5a_{5} + 6a_{6} + 7a_{7} + 8a_{8} + 9a_{9} + 10a_{10} + ...)x +\\ (a_{2} + 3a_{3} + 6a_{4} + 10a_{5} + 15a_{6} + 21a_{7} + 28a_{8} + 36a_{9} + 45a_{10} + ...)x^2 +\\ (a_{3} + 4a_{4} + 10a_{5} + 20a_{6} + 35a_{7} + 56a_{8} + 84a_{9} + 120a_{10} + ...)x^3...$ It makes a super nice linear system, with determinant=1. I bet that it is well studied and known. His inverse matrix is the same matrix, but each element has alternate sign. Here is the inverse of the coefficients matrix, generated with PariGP Code:gp > m=matrix(Size, Size,n,i,binomial(i-1,n-1)) (20:22) gp > m^-1 [1 -1  1 -1  1  -1   1  -1   1   -1    1   -1    1    -1     1    -1      1     -1      1     -1 [0  1 -2  3 -4   5  -6   7  -8    9  -10   11  -12    13   -14    15    -16     17    -18     19 [0  0  1 -3  6 -10  15 -21  28  -36   45  -55   66   -78    91  -105    120   -136    153   -171 [0  0  0  1 -4  10 -20  35 -56   84 -120  165 -220   286  -364   455   -560    680   -816    969 [0  0  0  0  1  -5  15 -35  70 -126  210 -330  495  -715  1001 -1365   1820  -2380   3060  -3876 [0  0  0  0  0   1  -6  21 -56  126 -252  462 -792  1287 -2002  3003  -4368   6188  -8568  11628 [0  0  0  0  0   0   1  -7  28  -84  210 -462  924 -1716  3003 -5005   8008 -12376  18564 -27132 [0  0  0  0  0   0   0   1  -8   36 -120  330 -792  1716 -3432  6435 -11440  19448 -31824  50388 [0  0  0  0  0   0   0   0   1   -9   45 -165  495 -1287  3003 -6435  12870 -24310  43758 -75582 [0  0  0  0  0   0   0   0   0    1  -10   55 -220   715 -2002  5005 -11440  24310 -48620  92378 [0  0  0  0  0   0   0   0   0    0    1  -11   66  -286  1001 -3003   8008 -19448  43758 -92378 [0  0  0  0  0   0   0   0   0    0    0    1  -12    78  -364  1365  -4368  12376 -31824  75582The inverse can be directly generated with m=matrix(Size, Size,n,i,(-1)^(n+i)*binomial(i-1,n-1)); if you want to test it with Sheldonison kneser.gp, use Size=60, and get the coefficients with ai=matrix(60,1,f,c,polcoeff(xsexp,f-1)). Now, let's deal with the right side of the first equation: This is the Taylor expansion for a^x: $a^x\,=\, \sum_{n=0}^{\infty}\frac{ln(a)^n}{n!}.x^n$ Let's compose with the Taylor polynomial at the start of the post: ${\color{Red} {a^{^xa}\,=\, \sum_{n=0}^{\infty}\frac{ln(a)^n}{n!}.{\left(\sum_{m=0}^{\infty}{a_m .x^m}\right)}^n$ (We should apply the multinomial theorem) I made an expansion of a^x with 12 terms, replaced each x with a polynomial with 12 terms: Code:(01:40) gp > exp(lna*x) %1 = 1 + lna*x + 1/2*lna^2*x^2 + 1/6*lna^3*x^3 + 1/24*lna^4*x^4 + 1/120*lna^5*x^5 + 1/720*lna^6*x^6 + 1/5040*lna^7*x^7 + 1/40320*lna^8*x^8 + 1/362880*lna^9*x^9 + 1/3628800*lna^10*x^10 + 1/39916800*lna^11*x^11 Code:1 + lna*(1+a1*x+a2*x^2+a3*x^3+a4*x^4+a5*x^5+a6*x^6+a7*x^7+a8*x^8+a9*x^9+a10*x^10+a11*x^11) +1/2*lna^2*(1+a1*x+a2*x^2+a3*x^3+a4*x^4+a5*x^5+a6*x^6+a7*x^7+a8*x^8+a9*x^9+a10*x^10+a11*x^11)^2 +1/6*lna^3*(1+a1*x+a2*x^2+a3*x^3+a4*x^4+a5*x^5+a6*x^6+a7*x^7+a8*x^8+a9*x^9+a10*x^10+a11*x^11)^3 +1/24*lna^4*(1+a1*x+a2*x^2+a3*x^3+a4*x^4+a5*x^5+a6*x^6+a7*x^7+a8*x^8+a9*x^9+a10*x^10+a11*x^11)^4 +1/120*lna^5*(1+a1*x+a2*x^2+a3*x^3+a4*x^4+a5*x^5+a6*x^6+a7*x^7+a8*x^8+a9*x^9+a10*x^10+a11*x^11)^5 +1/720*lna^6*(1+a1*x+a2*x^2+a3*x^3+a4*x^4+a5*x^5+a6*x^6+a7*x^7+a8*x^8+a9*x^9+a10*x^10+a11*x^11)^6 +1/5040*lna^7*(1+a1*x+a2*x^2+a3*x^3+a4*x^4+a5*x^5+a6*x^6+a7*x^7+a8*x^8+a9*x^9+a10*x^10+a11*x^11)^7 +1/40320*lna^8*(1+a1*x+a2*x^2+a3*x^3+a4*x^4+a5*x^5+a6*x^6+a7*x^7+a8*x^8+a9*x^9+a10*x^10+a11*x^11)^8 +1/362880*lna^9*(1+a1*x+a2*x^2+a3*x^3+a4*x^4+a5*x^5+a6*x^6+a7*x^7+a8*x^8+a9*x^9+a10*x^10+a11*x^11)^9 +1/3628800*lna^10*(1+a1*x+a2*x^2+a3*x^3+a4*x^4+a5*x^5+a6*x^6+a7*x^7+a8*x^8+a9*x^9+a10*x^10+a11*x^11)^10 +1/39916800*lna^11*(1+a1*x+a2*x^2+a3*x^3+a4*x^4+a5*x^5+a6*x^6+a7*x^7+a8*x^8+a9*x^9+a10*x^10+a11*x^11)^11 ...and threw it to PariGp to expand. I got this 24Mb text file http://s000.tinyupload.com/index.php?fil...4013545746 Since the series are truncated, I made some guess about what each term means: $a^{^xa}= \left(\sum_{n=0}^{...}{\frac{lna^n}{n!}} \right) \,+\, \left(a_1.\sum_{n=0}^{...}{\frac{lna^{n+1}}{n!}} \right).x \,+\, \left(a_2 .\sum_{n=0}^{...}{\frac{lna^{n+1}}{n!}} \,+\, a_1^2.\sum_{n=0}^{...}{\frac{lna^{2+n}}{2.n!}} \right).x^2\,+\, \\ \left(a_3 .\sum_{n=0}^{...}{\frac{lna^{n+1}}{n!}} \,+\, a_1.a_2.\sum_{n=0}^{...}{\frac{lna^{2+n}}{n!}} \,+\, a_1^3 .\sum_{n=0}^{...}{\frac{lna^{3+n}}{6.n!}} \right).x^3 \,+\, \left(a_4 .\sum_{n=0}^{...}{\frac{lna^{n+1}}{n!}} \,+\, a_2^2 .\sum_{n=0}^{...}{\frac{lna^{2+n}}{2.n!}} \,+\, a_3.a_1 .\sum_{n=0}^{...}{\frac{lna^{2+n}}{n!}} \,+\, a_2.a_1^2.\sum_{n=0}^{...}{\frac{lna^{3+n}}{2.n!}} \,+\, a_1^4 .\sum_{n=0}^{...}{\frac{lna^{4+n}}{24.n!}} \right).x^4 \,+\, ...$ ${\color{Red} { a^{^xa}= a \,+\, lna.a.a_1 \,.\, x \,+\, (lna.a.a_2 + \frac{lna^2.a}{2}.a_1^2) \,.\, x^2 \,+\, (lna.a.a_3 + lna^2.a.a_1.a_2 +\frac{lna^3.a}{6}.a_1^3) \,.\, x^3 \,+\, (lna.a.a_4 + lna^2.a.a_3.a_1 + \frac{lna^2.a}{2}.a_2^2 +\frac{lna^3.a}{2}.a_2.a_1^2+\frac{lna^4.a}{24}.a_1^4) \,.\,x^4 \\ +(lna.a.a_5 + lna^2.a.a_1.a_4 + {lna^2.a}.a_3.a_2 + \frac{lna^3.a}{2}.a_3.a_1^2 + \frac{lna^3.a}{2}.a_1.a_2^2+\frac{ lna^4.a}{6}.a_2.a_1^3+\frac{lna^5.a}{120}.a_1^5) \,.\,x^5 \\ +(lna.a.a_6 + {lna^2.a}.a_1.a_5 + lna^2.a.a_2.a_4 + \frac{lna^3.a}{2}.a_4.a_1^2 + \frac{lna^2.a}{2}.a_3^2 + { lna^3.a}.a_1.a_3.a_2 + \frac{lna^3.a}{6}.a_2^3 +\frac{lna^4.a}{4}.a_2^2 . a_1^2+\frac{lna^4.a}{6}.a_3.a_1^3+\frac{lna^5.a}{24}.a_2.a_1^4+\frac{lna^6.a}{ 720}.a_1^6) \,.\,x^6 \\ }$ ${\color{Red} { +a.( lna.a_7 + lna^2. a_4.a_3 + lna^2.a_5.a_2 + \frac{lna^3}{2}a_3.a_2^2 + lna^2.a_6.a_1 + \frac{lna^3}{2}a_3^2.a_1 + lna^3a_4.a_2.a_1 + \frac{lna^4}{6}a_2^3.a_1 + \frac{lna^3}{2}a_5.a_1^2 + \frac{lna^4}{2} a_3.a_2.a_1^2 + \frac{lna^4}{6}a_4.a_1^3 + \frac{lna^5}{12} a_2^2.a_1^3 + \frac{lna^5}{24}a_3.a_1^4 + \frac{lna^6}{120}a_2.a_1^5 + \frac{lna^7}{5040}.a_1^7 )\,.\,x^7 }$ EDIT: more terms: ${\color{Red} { +a.(ln.a.a_8 + \frac{lna^2}{2} .a_4^2 + lna^2 .a_3.a_5 + lna^2 .a_2 .a_6 + \frac{lna^3}{2} .a_2 .a_3^2 + \frac{lna^3}{2} .a_2^2 .a_4 + \frac{lna^4}{24} .a_2^4 + lna^2 .a_1 .a_7 + lna^3 .a_1 .a_3 .a_4 + lna^3 .a_1 .a_2 .a_5 + \frac{lna^4}{2} .a_1 .a_2^2 .a_3 + \frac{lna^3}{2} .a_1^2 .a_6 + \\ \frac{lna^4}{4} .a_1^2 .a_3^2 + \frac{lna^4}{2} .a_1^2 .a_2 .a_4 + \frac{lna^5}{12} .a_1^2 .a_2^3 + \frac{lna^4}{6}.a_1^3 .a_5 + \frac{lna^5}{6} .a_1^3 .a_2 .a_3 + \frac{lna^5}{24} .a_1^4 .a_4 + \frac{lna^6}{48} .a_1^4 .a_2^2 + \frac{lna^6}{120} .a_1^5 .a_3 + \frac{lna^7}{720} .a_1^6 .a_2 + \frac{lna^8}{40320} .a_1^8 )\,.\,x^8 }$ ${\color{Red} { +a.(lna.a_9 + lna^2.a_4 .a_5 + lna^2 .a_3 .a_6 + \frac{lna^3}{6} .a_3^3 + lna^2.a_2 .a_7 + lna^3.a_2 .a_3 .a_4 + \frac{lna^3}{2}.a_2^2 .a_5 + \frac{lna^4}{6} .a_2^3 .a_3 + lna^2.a_1 .a_8 + \frac{lna^3}{2} .a_1 .a_4^2 + \\ lna^3 .a_1 .a_3 .a_5 + lna^3 .a_1 .a_2 .a_6 + \frac{lna^4}{2} .a_1 .a_2 .a_3^2 + \frac{lna^4}{2} .a_1 .a_2^2 .a_4 + \frac{lna^5}{24}.a_1 .a_2^4 + \frac{lna^3}{2}.a_1^2 .a_7 + \frac{lna^4}{2}.a_1^2 .a_3 .a_4 + \frac{lna^4}{2} .a_1^2 .a_2 .a_5 + \frac{lna^5}{4}.a_1^2 .a_2^2 .a_3 + \frac{lna^4}{6} .a_1^3 .a_6 +\\ \frac{lna^5}{12} .a_1^3 .a_3^2 + \frac{lna^5}{6} .a_1^3 .a_2 .a_4 + \frac{lna^6}{36}.a_1^3 .a_2^3 + \frac{lna^5}{24}.a_1^4 .a_5 + \frac{lna^6}{24}.a_1^4 .a_2 .a_3 + \frac{lna^6}{120}.a_1^5 .a_4 + \frac{lna^7}{240} .a_1^5 .a_2^2 + \frac{lna^7}{720} .a_1^6 .a_3 + \frac{lna^8}{5040}.a_1^7 .a_2 + \frac{lna^9}{362880}.a_1^9 ).x^9 }$ This last expression seems to be related to the Taylor expansion of $\vspace{15}{^xa}$ around x=1. So, each summatory of coefficients, for each power of x, is in some loose sense, "the derivative" of the summatory of the precedent power of x (taking da/dx=lna.a.a₁, and daᵢ/dx=aᵢ₊₁) I wrote "in some loose sense", because there are multiplicative constants missing. For example, the "derivative" of lna.a.a₁ "should be" (lna.a.a₂+lna².a.a₁²), but instead is (lna.a.a₂+lna².a.a₁²/n!), where n is the power exponent of a₁ in that term. again the "derivative" of (lna.a.a₂+lna².a.a₁²/2!) "should be" (lna.a.a₃ + 2 . lna².a.a₁.a₂ +lna³.a.a₁³/2), but instead is (lna.a.a₃ + lna².a.a₁.a₂ +lna³.a.a₁³/3!) ¿Can you find the right rule for the coefficients of each power of x? Here is an excel worksheet, which takes the coefficients calculated with Sheldonison Kneser.gp program (for base a=Pi), and compares the blue and red equations: http://s000.tinyupload.com/index.php?fil...7813385010 Unfortunately, this website does not allow to upload most format files, so I uploaded it o Tinyupload.com, but there is no guarantee of permanence, so it may be deleted in the future. I have the result, but I do not yet know how to get it. marraco Fellow Posts: 93 Threads: 11 Joined: Apr 2011 05/01/2015, 01:57 AM (This post was last modified: 05/01/2015, 11:11 AM by marraco.) This is a numerical example for base a=e I had shown how to get 2 sides of a system of equations for solving the coefficients. One side is blue (the linear one), and the other is red (nonlinear, and I still don't know the pattern to generate the complete system of equations. I only got up to x^7, and I can get more, but of course, what I need is a more general expression. For base e, you get a system of equations. you get this blue side for 7 coefficients (it is independent of the base): Code:(21:23) gp > Size=9 %3 = 9 (21:24) gp > m=matrix(Size, Size,n,i,binomial(i-1,n-1)) %4 = [1 1 1 1 1  1  1  1  1] [0 1 2 3 4  5  6  7  8] [0 0 1 3 6 10 15 21 28] [0 0 0 1 4 10 20 35 56] [0 0 0 0 1  5 15 35 70] [0 0 0 0 0  1  6 21 56] [0 0 0 0 0  0  1  7 28] [0 0 0 0 0  0  0  1  8] [0 0 0 0 0  0  0  0  1] and this is the red side: ${\color{Red} { e^{^xe}= e \,+\, e.a_1 \,.\, x \,+\, (e.a_2 + \frac{e}{2}.a_1^2) \,.\, x^2 \,+\, (e.a_3 + e.a_1.a_2 +\frac{e}{6}.a_1^3) \,.\, x^3 \,+\, (e.a_4 + \frac{e}{2}.a_2^2 +e.a_3.a_1+\frac{e}{2}.a_2.a_1^2+\frac{e}{24}.a_1^4) \,.\,x^4 \\ +(e.a_5 + {e}.a_3.a_2 + e.a_1.a_4+\frac{e}{2}.a_1.a_2^2+\frac{e}{2}.a_3.a_1^2+\frac{e}{6}.a_2.a_1^3+ \frac{e}{120} .a_1^5) \,.\,x^5 \\ +(e.a_6 + \frac{e}{2}.a_3^2 + e.a_2.a_4+\frac{e}{6}.a_2^3+ {e}.a_1.a_5+ {e}.a_1.a_3.a_2+\frac{e}{2}.a_4.a_1^2+\frac{e}{4}.a_2^2 . a_1^2+\frac{e}{6}.a_3.a_1^3+\frac{e}{24}.a_2.a_1^4+\frac{e}{ 720}.a_1^6) \,.\,x^6 \\ }$ ${\color{Red} { +e.(a_7 + 1^2. a_4.a_3 + a_5.a_2 + \frac{1}{2}a_3.a_2^2 + a_6.a_1 + \frac{1}{2}a_3^2.a_1 + a_4.a_2.a_1 + \frac{1}{6}a_2^3.a_1 + \frac{1}{2}a_5.a_1^2 + \frac{1}{2} a_3.a_2.a_1^2 + \frac{1}{6}a_4.a_1^3 + \frac{1}{12} a_2^2.a_1^3 + \frac{1}{24}a_3.a_1^4 + \frac{1}{120}a_2.a_1^5 + \frac{1}{5040}.a_1^7 )\,.\,x^7 }$ So you get this systems of equations (blue to the left, and red to the right): Code:[1 1 1 1 1  1  1  1  1]    [ 1]   [e] [0 1 2 3 4  5  6  7  8]    [a₁]   [e.a₁] [0 0 1 3 6 10 15 21 28]    [a₂]   [e.a₂+e/2.a₁²] [0 0 0 1 4 10 20 35 56]    [a₃]   [e.a₃+e.a₁.a₂+e/6.a₁³] [0 0 0 0 1  5 15 35 70]  * [a₄] = [e.a₄+e/2.a₂²+e.a₃.a₁+e/2.a₂.a1²+e/24.a₁⁴] [0 0 0 0 0  1  6 21 56]    [a₅]   [...] [0 0 0 0 0  0  1  7 28]    [a₆]   [...] [0 0 0 0 0  0  0  1  8]    [a₇]   [...] [0 0 0 0 0  0  0  0  1]    [a₈]   [...] It is a non linear system of equations, and the solution for this particular case is: a₀= 1,00000000000000000 a₁= 1,09975111049169000 a₂= 0,24752638354178700 a₃= 0,15046151104294100 a₄= 0,12170896032120000 a₅= 0,16084324512292400 a₆= -0,02254254634348470 a₇= -0,10318144159688800 a₈= 0,06371479195361670 Compare with Sheldonison's kneser.gp solution: Code:a0=   1.000000000000000000000000000000000000000000000000000000000000000 a1=   1.091767351258320991801384550027151644384731177193748032852736750 a2=   0.2714832129016945953317066836235490061739872161990598817820694029 a3=   0.2124532481762562843089676377409482685666305767555868534429427250 a4=   0.06954037613998737372867423270746944766611954537384154515961000453 a5=   0.04429195209047330440644034438551461331201912452396122603445409319 a6=   0.01473674209638939115209628691553407087036283284802524310071901416 a7=   0.008668781817225260366380392529639960019193562974233192587504340111 a8=   0.002796479398385459694825991301149597269227881323260081302040577606 a9=   0.001610631290584272072162645164026110049792896861436056957822790182 a10=  0.0004899272314843773346986672258324787189179236966986597429129832332 a11=  0.0002881810711540458113452640412964722721968311050442105340877181595 a12=  8.009461253854333344427358300999263618841918537261023130790163984 E-5 a13=  5.029114179380540369459011462420351199253676250298573251465345775 E-5 a14=  1.218379034490009161619171109859335192564698424745403501454238589 E-5 a15=  8.665533667381574685245804554105291695042724874461047071879364312 E-6 a16=  1.687782319317538991789009317583785512852144820496380916395175482 E-6 a17=  1.493253248573492581066504431732751485456118359869377142782801855 E-6 a18=  1.987607642049274553198189794968196417348167228423221823766208771 E-7 a19=  2.608673560043263731645821608532933364457829738518570518951024854 E-7 a20=  1.470995414254190186141218818247639464249470033298457423356618453 E-8 a21=  4.683449732741350625509370993006576850859342869408936466444526422 E-8 a22= -1.549241665546769521805465176448252112209562873661024180269440066 E-9 a23=  8.741510781350935912992558117122349583609707251224417742494147299 E-9 a24= -1.125787310103062317575134515738378022835159203164294437887008675 E-9 a25=  1.707959267270728412565608778729726960430408931538582375779659881 E-9 a26= -3.778583154922985176492143492500296036058409993500345462555638121 E-10 a27=  3.495778765110216317873145649935531461006488191616787742019992521 E-10 a28= -1.053770123445001506629425792917061970255456556066414419570018689 E-10 a29=  7.459097147607505280732283202189701957070179874063753558099592924 E-11 a30= -2.717598206577734869329877172492724299204938962536131010671806592 E-11 a31=  1.646076610661447130388508182175742339593708736967841194185926886 E-11 a32= -6.741873152405052999147453463677312005415532501572568628993007553 E-12 a33=  3.725328723319468544317086960688595471124134410671562760851241992 E-12 a34= -1.639087326793590223458207893420015494534981157084934821663605376 E-12 a35=  8.583638311358568060488665543255939724011799192373411667290837403 E-13 a36= -3.943738739105384313579489883439749145625771519735734916959207765 E-13 a37=  2.002523128021887055893526704591127887891070857821009841312866823 E-13 a38= -9.441962242924065023715111579990954292712006432497954997730218600 E-14 a39=  4.712054745849371340817414393412613311684787369946584046802094568 E-14 a40= -2.256291882035597080043272706155245325625189406689252333379604487 E-14 a41=  1.115468850616536996293093710617388942901924987321949988173687870 E-14 a42= -5.390745557016350491840931638954438791176540897408393713471988968 E-15 a43=  2.652158491516681872817207767927177972525297352287457523410538087 E-15 a44= -1.288910765544553681933994492897094530731798239089453308045916388 E-15 a45=  6.326678501956660453007840302627916831673424077807564985330246473 E-16 a46= -3.085457150492335988961833456292213681855982339566944345179362092 E-16 a47=  1.513176771782740527337006888790219839276363051364244735670911612 E-16 a48= -7.396534137094751433579658688811828771999777094874553108587585999 E-17 a49=  3.626987671054187604858900781140139207944144420079830647639126832 E-17 a50= -1.775725598676298403622157433145771025580236336491145640801940287 E-17 a51=  8.709879544396054644216187847273723532343829279009721890829235905 E-18 a52= -4.269289282339156322311235369711703332985224326874010775539646161 E-18 a53=  2.095044162575528110270707271710047732715299297962567107013275560 E-18 a54= -1.027883709282258792101512551665977860702273246084671587434733006 E-18 a55=  5.046824247438176461629463228342973276669090061957443080345413943 E-19 a56= -2.478050595821552304759308268057851431159467131543548274881868429 E-19 a57=  1.217394203039331646009323993543125208245611419932029287538356629 E-19 a58= -5.981648632303782947485566789252901875389559922820136772179247709 E-20 a59=  2.940264344513896261748193069655017265616156575476060556667439589 E-20 This is a comparison of the coefficients obtained from both methods. Of course I only calculated (7+1dummy) coefficients, with excel precision, and Sheldonison code did 60 coefficients with 134 bits of precision. (I estimate that I need at least between 20 and 30 coefficients, and have no clue on how to solve it). The advantage of this method is that it does not depend of the base, so it will be able to solve negative bases. I have the result, but I do not yet know how to get it. tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 05/01/2015, 08:37 AM Im not completely sure about the coëfficiënts , but I can prove this converges as a function to sexp(x) !!! Not sure how to prove it is analytic though. Regards Tommy1729 marraco Fellow Posts: 93 Threads: 11 Joined: Apr 2011 05/01/2015, 09:42 AM (05/01/2015, 08:37 AM)tommy1729 Wrote: Im not completely sure about the coëfficiënts , but I can prove this converges as a function to sexp(x) !!! Please, do it! I have the result, but I do not yet know how to get it. sheldonison Long Time Fellow Posts: 631 Threads: 22 Joined: Oct 2008 05/01/2015, 01:04 PM (05/01/2015, 01:57 AM)marraco Wrote: This is a numerical example for base a=e I had shown how to get 2 sides of a system of equations for solving the coefficients. One side is blue (the linear one), and the other is red (nonlinear, and I still don't know the pattern to generate the complete system of equations. I only got up to x^7, and I can get more, but of course, what I need is a more general expression. ....You might want to compare this to the intuitive slog method. Jay took that farther than anyone else, getting results accurate to 21 decimal digits using his accelerated technique. http://math.eretrandre.org/tetrationforu...php?tid=63 With 8 terms, accuracy is limited to magnitude of the eighth term, and your eighth term is actually larger than Kneser's. If you were able to generate a 20 term approximation, then theoretically, one could have a around 14 decimal digits of accuracy ... assuming the terms scale based on the nearest logarithmic singularity at sexp(-2). For base e, the Taylor's series coefficients of Kneser's solution approach log(x+2) as n gets larger. - Sheldon tommy1729 Ultimate Fellow Posts: 1,358 Threads: 330 Joined: Feb 2009 05/01/2015, 09:43 PM Assuming a_n Goes to < (2/3)^n ; ( this gives us a sufficiently Large radius such that the equation is satisfied within the ROC.) Taylors theorem gives us f(x+1) = f(x) + f ' (x) + f " (x)/2 + ... Hence what the truncation of degree k solves locally is near ; f(x+1) + O(a_k x^k) = exp(f(x)) f(0)=1 By taking k Large and x small we get : f(0)=1 f(x+1)=exp(f(x)) + o(f(1)). ( take x < 1 to see this ) Notice lim o(f(1)) = lim a_k = 0. Hence we have in the limit k to oo assuming the ROC ; f(0)=1 f(x+1) = exp(f(x)) Qed So the attention Goes completely to the asymp of a_n. Hope that is clear. Q: Can we show existance and uniqueness for these equations FORMALLY ? Q2 ; i Will post in a new thread. Regards Tommy1729 marraco Fellow Posts: 93 Threads: 11 Joined: Apr 2011 05/03/2015, 04:35 AM (This post was last modified: 05/13/2015, 02:43 PM by marraco.) ${\color{Red} { a^{^xa}=\\ \,\,\,\,\,\,\,\,\,\,\,\,a \\ \,+\, lna.a.a_1 \,.\, x \\ \,+\, (lna.a.a_2 + \frac{lna^2}{2}.a.a_1^2) \,.\, x^2 \\ \,+\, (lna.a.a_3 + lna^2.a.a_1.a_2 +\frac{lna^3.a}{6}.a_1^3) \,.\, x^3 \\ \,+\, (lna.a.a_4 + lna^2.a.a_1.a_3 + \frac{lna^2.a}{2}.a_2^2 +\frac{lna^3.a}{2}.a_2.a_1^2+\frac{lna^4.a}{24}.a_1^4) \,.\,x^4 \\ \,+\, (lna.a.a_5 + lna^2.a.a_1.a_4 + {lna^2.a}.a_2.a_3 + \frac{lna^3.a}{2}.a_3.a_1^2 +\frac{ lna^4.a}{6}.a_2.a_1^3+ \frac{lna^3.a}{2}.a_1.a_2^2+\frac{lna^5.a}{120}.a_1^5) \,.\,x^5 \\ \,+\, (lna.a.a_6 + {lna^2.a}.a_1.a_5 + lna^2.a.a_2.a_4 + \frac{lna^3.a}{2}.a_4.a_1^2 +\frac{lna^4.a}{6}.a_3.a_1^3+ \frac{lna^2.a}{2}.a_3^2 + { lna^3.a}.a_1.a_3.a_2 + \frac{lna^3.a}{6}.a_2^3 +\frac{lna^4.a}{4}.a_2^2 . a_1^2+\frac{lna^5.a}{24}.a_2.a_1^4+\frac{lna^6.a}{ 720}.a_1^6) \,.\,x^6 \\ }$ $\mathbf{ -The \,\ number \,\ of \,\ terms \,\ for \,\ each \,\ element \,\ in \,\ the \,\ sequence }$ This is the number of terms from the first to the -power of x⁹- element: 1, 1, 2, 3, 5, 7, 11, 15, 22, 30 I threw the sequence to The On-Line Encyclopedia of Integer Sequences http://oeis.org/search?q=1%2C2%2C3%2C5%2...&go=Search and I got many different ways to calculate it. The first one is: "number of partitions of n (the partition numbers)." which PariGP can calculate with the numbpart() function: gp > a=vector(20,k,numbpart(k)) %3 = [1, 2, 3, 5, 7, 11, 15, 22, 30, 42, 56, 77, 101, 135, 176, 231, 297, 385, 490, 627] it works up to 56, but I truncated the series, so it should not work beyond that. $\mathbf{ -The \,\ index \,\ of \,\ the \,\ coefficients \,\ seems \,\ to \,\ be \,\ also \,\ the \,\ numbers \,\ on \,\ which \,\ n \,\ can \,\ be \,\ partitioned. }$ For example, the 4th element, for x⁴, has 5 elements, because 4 can be partitioned in 5 different ways. For each partition, there is an index i for each number in that partition: $ (lna.a.a_4 + lna^2.a.a_3.a_1 + \frac{lna^2.a}{2}.a_2^2 +\frac{lna^3.a}{2}.a_2.a_1^2+\frac{lna^4.a}{24}.a_1^4) \,.\,x^4 \\ \\ 4 \right a_4\\ 3 + 1 \right a_3.a_1\\ 2 + 2 \right a_2^2\\ 2 + 1 + 1 \right a_2.a_1^2\\ 1 + 1 + 1 + 1 \right a_1^4\\$ $\mathbf{ -The \,\ exponent \,\ of \,\ the \,\ logarithm \,\ lna \,\ is \,\ the \,\ quantity \,\ of \,\ elements \,\ on \,\ each \,\ partition: }$ $ (lna.a.a_4 + lna^2.a.a_3.a_1 + \frac{lna^2.a}{2}.a_2^2 +\frac{lna^3.a}{2}.a_2.a_1^2+\frac{lna^4.a}{24}.a_1^4) \,.\,x^4 $ $ 4 \right 1 \, elements \right lna \\ 3 + 1 \right 2 \,elements \right lna^2 \\ 2 + 2 \right 2 \, elements \right lna^2 \\ 2 + 1 + 1 \right 3 \, elements \right lna^3 \\ 1 + 1 + 1 + 1 \right 4 \, elements \right lna^4 \\$ $\mathbf{ -The \,\ integer \,\ divisor \,\ is \,\ the \,\ product \,\ of \,\ the \,\ factorials \,\ of \,\ the \,\ exponents \,\ of \,\ a_i }$ $ +a.(lna.a_9 + lna^2.a_4 .a_5 + lna^2 .a_3 .a_6 + \frac{lna^3}{6} .a_3^3 + lna^2.a_2 .a_7 + lna^3.a_2 .a_3 .a_4 + \frac{lna^3}{2}.a_2^2 .a_5 + \frac{lna^4}{6} .a_2^3 .a_3 + lna^2.a_1 .a_8 + \frac{lna^3}{2} .a_1 .a_4^2 + \\ lna^3 .a_1 .a_3 .a_5 + lna^3 .a_1 .a_2 .a_6 + \frac{lna^4}{2} .a_1 .a_2 .a_3^2 + \frac{lna^4}{2} .a_1 .a_2^2 .a_4 + \frac{lna^5}{24}.a_1 .a_2^4 + \frac{lna^3}{2}.a_1^2 .a_7 + \frac{lna^4}{2}.a_1^2 .a_3 .a_4 + \frac{lna^4}{2} .a_1^2 .a_2 .a_5 + \frac{lna^5}{4}.a_1^2 .a_2^2 .a_3 + \frac{lna^4}{6} .a_1^3 .a_6 +\\ \frac{lna^5}{12} .a_1^3 .a_3^2 + \frac{lna^5}{6} .a_1^3 .a_2 .a_4 + \frac{lna^6}{36}.a_1^3 .a_2^3 + \frac{lna^5}{24}.a_1^4 .a_5 + \frac{lna^6}{24}.a_1^4 .a_2 .a_3 + \frac{lna^6}{120}.a_1^5 .a_4 + \frac{lna^7}{240} .a_1^5 .a_2^2 + \frac{lna^7}{720} .a_1^6 .a_3 + \frac{lna^8}{5040}.a_1^7 .a_2 + \frac{lna^9}{362880}.a_1^9 ).x^9 $ $\small { 9\,\,\right\,\, a_9^1\,\,\right\,\, 1! \,=\, 1 \\ 1,\, 8\,\,\right\,\,a_1^1\,.\,a_8^1\,\,\right\,\,1!.1! \,=\, 1 \\ 2,\, 7\,\,\right\,\, a_2^1\,.\,a_7^1\,\,\right\,\, 1!.1! \,=\, 1 \\ 3,\, 6\,\,\right\,\, a_3^1\,.\,a_6^1\,\,\right\,\, 1!.1! \,=\, 1 \\ 4,\, 5\,\,\right\,\, a_4^1\,.\,a_5^1\,\,\right\,\, 1!.1! \,=\, 1 \\ 1,\, 1,\, 7\,\,\right\,\,a_1^2\,.\,a_7^1\,\,\right\,\,2!.1! \,=\, 2 \\ 1,\, 2,\, 6\,\,\right\,\,a_1^1\,.\,a_2^1\,.\,a_6^1\,\,\right\,\,1!.1!.1! \,=\, 1 \\ 1,\, 3,\, 5\,\,\right\,\,a_1^1\,.\,a_3^1\,.\,a_5^1\,\,\right\,\,1!.1!.1! \,=\, 1 \\ 1,\, 4,\, 4\,\,\right\,\,a_1^1\,.\,a_4^2\,\,\right\,\,1!.2! \,=\, 2 \\ 2,\, 2,\, 5\,\,\right\,\, a_2^2\,.\,a_5^1\,\,\right\,\, 2!.1! \,=\, 2 \\ 2,\, 3,\, 4\,\,\right\,\, a_2^1\,.\,a_3^1\,.\,a_4^1\,\,\right\,\, 1!.1!.1! \,=\, 1 \\ 3,\, 3,\, 3\,\,\right\,\, a_3^3\,\,\right\,\, 3! \,=\, 6 \\ 1,\, 1,\, 1,\, 6\,\,\right\,\,a_1^3\,.\,a_6^1\,\,\right\,\,3!.1! \,=\, 6 \\ 1,\, 1,\, 2,\, 5\,\,\right\,\,a_1^2\,.\,a_2^1\,.\,a_5^1\,\,\right\,\,2!.1!.1! \,=\, 2 \\ 1,\, 1,\, 3,\, 4\,\,\right\,\,a_1^2\,.\,a_3^1\,.\,a_4^1\,\,\right\,\,2!.1!.1! \,=\, 2 \\ 1,\, 2,\, 2,\, 4\,\,\right\,\,a_1^1\,.\,a_2^2\,.\,a_4^1\,\,\right\,\,1!.2!.1! \,=\, 2 \\ 1,\, 2,\, 3,\, 3\,\,\right\,\,a_1^1\,.\,a_2^1\,.\,a_3^2\,\,\right\,\,1!.1!.2! \,=\, 2 \\ 2,\, 2,\, 2,\, 3\,\,\right\,\, a_2^3\,.\,a_3^1\,\,\right\,\, 3!.1! \,=\, 6 \\ 1,\, 1,\,1,\, 1,\, 5\,\,\right\,\,a_1^4\,.\,a_5^1\,\,\right\,\,4!.1! \,=\, 24 \\ 1,\, 1,\, 1,\, 2,\, 4\,\,\right\,\,a_1^3\,.\,a_2^1\,.\,a_4^1\,\,\right\,\,3!.1!.1! \,=\, 6 \\ 1,\, 1,\, 1,\, 3,\, 3\,\,\right\,\,a_1^3\,.\,a_3^2\,\,\right\,\,3!.2! \,=\, 12 \\ 1,\, 1,\, 2,\, 2,\, 3\,\,\right\,\,a_1^2\,.\,a_2^2\,.\,a_3^1\,\,\right\,\,2!.2!.1! \,=\, 4 \\ 1,\, 2,\, 2,\, 2,\, 2\,\,\right\,\,a_1^1\,.\,a_2^4\,\,\right\,\,1!.4! \,=\, 24 \\ 1,\, 1,\, 1,\, 1,\, 1,\, 4\,\,\right\,\,a_1^5\,.\,a_4^1\,\,\right\,\,5!.1! \,=\, 120 \\ 1,\, 1,\, 1,\, 1,\, 2,\, 3\,\,\right\,\,a_1^4\,.\,a_2^1\,.\,a_3^1\,\,\right\,\,4!.1!.1! \,=\, 24 \\ 1,\, 1,\, 1,\, 2,\, 2,\, 2\,\,\right\,\,a_1^3\,.\,a_2^3\,\,\right\,\,3!.3! \,=\, 36 \\ 1,\, 1,\, 1,\, 1,\, 1,\, 1,\, 3\,\,\right\,\,a_1^6\,.\,a_3^1\,\,\right\,\,6!.1! \,=\, 720 \\ 1,\, 1,\, 1,\, 1,\, 1,\, 2,\, 2\,\,\right\,\,a_1^5\,.\,a_2^2\,\,\right\,\,5!.2! \,=\, 240 \\ 1,\, 1,\, 1,\, 1,\, 1,\, 1,\, 1,\, 2\,\,\right\,\,a_1^7\,.\,a_2^1\,\,\right\,\,7!.1! \,=\, 5040 \\ 1,\, 1,\, 1,\, 1,\, 1,\, 1,\, 1,\, 1,\, 1\,\,\right\,\,a_1^9\,\,\right\,\,9! \,=\, 362880 \\ }$ I have the result, but I do not yet know how to get it. marraco Fellow Posts: 93 Threads: 11 Joined: Apr 2011 05/03/2015, 04:46 AM Of course, this is empirical. It should be rigorously deduced from the application of the multinomial theorem, but knowing the result should make the proof easier. I have the result, but I do not yet know how to get it. marraco Fellow Posts: 93 Threads: 11 Joined: Apr 2011 05/03/2015, 12:07 PM (This post was last modified: 05/03/2015, 12:09 PM by marraco.) This PariGP code generates the factors of X, exactly as written here: Code:\\ exponent of X for which the factor will be generated. expX=4 \\for each partition p, index i of a_i \\vector containing vectors. indexs=partitions(expX) num_partitions=length(indexs) \\exponent of lna lna_exp=vector(num_partitions,i,length(indexs[i])) \\exponent, in each partition, for each index of a_i \\Row=partition \\Column=exponent for ai index_exp=matrix(num_partitions,expX,r,c,sum(n=1,lna_exp[r],if(c-indexs[r][n],0,1))) \\integer divisor for each partition divisor=vector(num_partitions,p,truncate(prod(i=1,expX,factorial(index_exp[p,i])))) ai=[1,a1,a2,a3,a4,a5,a6,a7,a8,a9,a10,a11,a12] \\complete factor for the exponent X^expX Factor_for_Xp=sum(p=1,num_partitions,(a*lna^lna_exp[p])/divisor[p]*prod(i=1,expX,ai[i+1]^index_exp[p,i])) for example, for expX=4, it generates this output: 1/24*lna^4*a*a1^4 + 1/2*lna^3*a*a2*a1^2 + lna^2*a*a3*a1 + (1/2*lna^2*a*a2^2 + lna*a*a4) This probably is not appropriate for solving it with PariGP. I don't even know if PariGP is the appropriate tool for solving the equations. by the way, there is a procedure to programmatically generate this vector, for an arbitrary length? ai=[1,a1,a2,a3,a4,a5,a6,a7,a8,a9,a10,a11,a12] I have the result, but I do not yet know how to get it. Gottfried Ultimate Fellow Posts: 758 Threads: 117 Joined: Aug 2007 05/05/2015, 07:40 AM (This post was last modified: 05/05/2015, 08:17 AM by Gottfried.) Hmm, coming a bit late... However - this fiddling reminds me nicely of my own struggles, when I started to look at tetration. So I'll do the proposal to introduce matrix-notation, and especially that of Carleman-matrices, into that study. It has also the advantage, that the formulae can then immediately be (numerically approximate) checked by use of the matrix-features in Pari/GP. What I introduced first was the notation of a "vandermondevector" V(x) of ideally infinite, practically truncated size to size =n (and using that Vector- and matrixfunctions implicitely by the global variable "n") . In Pari/GP V(x,dim=n) = vector(dim,r,x^(r-1)) Then with a vector A_1, which contains the coefficients a0,a1,... a_{n-1} of some powerseries one can write down easily the approximate formula f(x) = V(x) * A_1 ~ \\ approximate due to truncation to "n" terms This shall already indicate your initial definition of the A-coefficients, and of course, A shall be so that f(x) = b^^x = V(x) * A_1~ If one needs also the powers of f(x) and thus some manipulations on A_1 (as in your post) one does with advantage using a Carlemanmatrix A, which is the concatenation of columnvectors, such that for some function f(x) V(x) * A = [1,f(x),f(x)^2, f(x)^3,... ] = V(f(x)) and again this is ideally of infinite columns but practically (in Pari/GP) truncated to n columns. With your idea of some coefficients a_k and now the matrix A instead, A is searched such that V(x) * A = [1, b^^x , (b^^x)^2 , (b^^ x)^3, ...] = V(b^^x) (I took the liberty to rename the base variable for the exponentiation to "b" (from "base") and to avoid confusion with the capital letter A and the small letters "a" for its entries) . Also I have a standardmatrix "Bb" which is a Carlemanmatrix for the exponentiation V(x) * Bb = V(b^x) beta = log(b) Bb = matrix(n,n,r,c, (beta*(c-1))^(r-1)/(r-1)!) With size n=32 and b=sqrt(2) it is easy to see the map x -> b^x with the visible digits (I do by default 200 dec digits internal float precision and 12 digits to show) Your convolution-formula is then simply written V(x) * A * Bb = V(b^^x) * Bb = V(b^(b^^x)) and the initial idea to be used is much concisely expressed in my notation here: V(x+1)*A = V(x)*A*Bb \\ searched for A, such that this equation holds You've done even more: you've introduced the binomial-rules, and -in principle- discovered, that the matrix of binomial-coefficients is of importance here. I call this matrix usually P, and here it should be used as upper triangular Pascalmatrix, such that (with its property being a Carlemanmatrix as well!) (in Pari/GP: P=matpascal(n-1) ~ ) V(x) * P = V(x+1) V(x) * P^h = V(x+h) \\ where this must be implemented programmatically in Pari/GP for fractional h We arrive now at the conditions of your first post: Thus your goal is: let us find a carlemanmatrix A such that V(x)*P*A = V(x)*A*Bb As long as the involved dot-products are convergent, there is a theorem, that if for a continuous variable x there is a nonzero interval, where V(x)*A = V(x)*B (for infinite size, thus for a power series), then A=B. The existence of this condition is important here, because (only) if A is of the required form, then also without the leading V(x)-vector it must be equal: P*A = A*Bb And now we can go on using knowledge from matrix-operations to try to solve this euqation for A. It reminds of an eigenvalue-system, however neither P nor Bb are diagonal here. I'm not yet stepping further here, I've limited time this weeks, but I'll come back later to this. Just for a short approximation of Sheldon's coefficients: Fractional iteration can be approximated by fractional powers of Bb: V(x) * Bb^h = V(exp_b^h(x)) V(1) * Bb^h = V(exp_b^h(1)) = V(b^^h) and fractional powers of Bb can be generated if Bb is diagonalizable. Now, for the approximation by the finite truncation, say n=16 and size = nxn Pari/GP can do this in a blink: tmpM=mateigen(Bb) tmpW=tmpM^-1 tmpD= diag(tmpW * Bb * tmpM ) and check, that indeed tmpM*matdiagonal(tmpD)*tmpW - Bb \\ approximately zero Fractional powers of Bb can now be constructed by fractional powers of the entries of tmpD only! For this I have also a standard-procedure: dpow(D,ex=1)= matdiagonal(vector(#D,r,D[r]^ex)) so tmpM*dpow(tmpD,h)*tmpW \\ is equal to Bb^h Because we want V(1)* B^x \\ as a powerseries in x I just did with the symbolic variable x and powerseries expansion to 32 terms (by setting \ps 32) V(1) * tmpM * dpow(tmpD,x) * tmpW[,2] to get the powerseries, with coefficients very near to Sheldon's coefficients. Unfortunately, for sizes nxn=32x32 the mateigen-procedure needs high precision (perhaps 800 digits or more) and the 64x64 version I could only compute with precision of 3200 internal dec digits, so this method is only for toying around and for the intuition. But one can see a comparision to Sheldons solution to base b=4 in my "comparision of methods" -(e-) paper which I've linked to in older postings here. V(1) ~ * (tmpM*(dpow(tmpD,x)*tmpW[,2] )) %910 = 1.00000000000 + 1.09176762501*x + 0.271483635676*x^2 + 0.212452965751*x^3 + 0.0695402029399*x^4 + 0.0442918971124*x^5 + 0.0147365956688*x^6 + 0.00866880451890*x^7 + 0.00279641322771*x^8 + 0.00161065399837*x^9 + O(x^10) (I'll continue this another day... and wish so far much joy and success on your detective-journey for the composition of power series!) Gottfried Gottfried Gottfried Helms, Kassel « Next Oldest | Next Newest »

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